Chapter 17, Problem 46CE

(a)

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with specification limits of 120 mg and 160 mg, process mean is 140 mg and the process standard deviation is 5 mg.

Answer to Problem 46CE

The value of $C_p$ is 1.33.

The value of $C_{pk}$ is 1.33.

Explanation of Solution

Calculation:

The given information is that, lower specification limit is 120 mg, upper specification limit is 160 mg, process mean is 140 mg and the process standard deviation is 5 mg.

The formula for $C_p$ is,

$C_p=\frac{USL-LSL}{6\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation.

The formula for $C_{pk}$ is,

$C_{pk}=\frac{\min \left(\mu -LSL,USL-\mu \right)}{3\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation, $\mu$ is the center line.

$C_p$:

Substitute, $\text{LSL}=120,\text{USL}=160,\sigma =5,\mu =140$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{160-120}{6\left(5\right)}\\ =\frac{40}{30}\\ =1.33\end{array}$

Hence, the value of $C_p$ is 1.33.

$C_{pk}$:

Substitute, $\text{LSL}=120,\text{USL}=160,\sigma =5,\mu =140$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(140-120,160-140\right)}{3\left(5\right)}\\ =\frac{\min \left(20,20\right)}{15}\\ =\frac{20}{15}\\ =1.33\end{array}$

Hence, the value of $C_{pk}$ is 1.33.

(b)

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with specification limits of 120 mg and 160 mg, process mean is 140 mg and the process standard deviation is 3 mg.

Answer to Problem 46CE

The value of $C_p$ is 2.22.

The value of $C_{pk}$ is 2.22.

Explanation of Solution

Calculation:

The given information is that, lower specification limit is 120 mg, upper specification limit is 160 mg, process mean is 140 mg and the process standard deviation is 3 mg.

The formula for $C_p$ is,

$C_p=\frac{USL-LSL}{6\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation.

The formula for $C_{pk}$ is,

$C_{pk}=\frac{\min \left(\mu -LSL,USL-\mu \right)}{3\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation, $\mu$ is the center line.

$C_p$:

Substitute, $\text{LSL}=120,\text{USL}=160,\sigma =3,\mu =140$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{160-120}{6\left(3\right)}\\ =\frac{40}{18}\\ =2.22\end{array}$

Hence, the value of $C_p$ is 2.22.

$C_{pk}$:

Substitute, $\text{LSL}=120,\text{USL}=160,\sigma =3,\mu =140$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(140-120,160-140\right)}{3\left(3\right)}\\ =\frac{\min \left(20,20\right)}{9}\\ =\frac{20}{9}\\ =2.22\end{array}$

Hence, the value of $C_{pk}$ is 2.22.